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 Complex Analysis Complex Analysis Math Forum

May 6th, 2019, 07:26 AM   #11
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 Originally Posted by Integrator Hello, I suppose we have to consider the possibility that $\displaystyle x = a + k(a)i$ where $\displaystyle a \in \mathbb R$ and $\displaystyle i^2=-1$ , which means that the initial equation will result an equation in the form $\displaystyle u + vi=0$ where $\displaystyle u=f(a)$ and $\displaystyle v=g(a)$.How do we solve the equation $\displaystyle f(a)+g(a)i=0$ where $\displaystyle i^2=-1$?Thank you very much! All the best, Integrator
Sorry, but I have no idea how one would find complex solutions for the initial equation you provided. Hopefully, someone else with the requisite knowledge can offer you some guidance. May 10th, 2019, 09:11 AM #12 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 6 Hello all, Solving the WolframAlpha computing program: https://www.wolframalpha.com/input/?...%2B2%7C%27%3D0 ----------------------------------------- The first two solutions given by "WolframAlpha" are the same as those found by me. From my calculations would also result solutions for $\displaystyle k(a)=0$ , $\displaystyle k(a)=i(a-2)$ , $\displaystyle k(a)=i(a-1)$ and $\displaystyle k(a)=\frac{i(2a-3)}{2}$ where $\displaystyle i^2=-1$ , $\displaystyle a=Re(x)$ and $\displaystyle k(a)=Im(x)$ , but I do not know how to interpret them to calculate $\displaystyle x=a+k(a)i$ .... All the best, Integrator Last edited by Integrator; May 10th, 2019 at 09:34 AM. Tags equation, modules, modulus Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post NAC54321 Physics 0 November 13th, 2018 05:03 AM bilano99 Calculus 2 March 15th, 2013 12:48 PM dpwheelwright Number Theory 5 August 5th, 2012 01:36 PM omega Complex Analysis 2 December 17th, 2011 02:51 PM Hemi08 Abstract Algebra 1 August 20th, 2009 07:54 PM

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